A man is known to speak the truth 2 out of 3 times. He throws a die and reports that the number obtained is a four. Find the probability that the number obtained is actually a four by using bayes theorem.

E1 be the event that four is obtained and E2 be its complementary event.

Probability that four occurs

P(E1) = 1/6

probability that four does not occur

P(E2) = 1- P(E1) = 1 – \$\frac(1)(6)\$ = \$\frac(5)(6)\$

probability that man reports four and it is actually a four

P(A/E1) = \$\frac(2)(3)\$

probability that man reports four and it is not a four2

P(A/E2) = \$\frac(1)(3)\$

Bayes theorem \$(P(E_1/A) = P(E_1) * P(A/E_1)) / (P(E_1) * P(A/E_1) + P(E_2) * P(A/E_2))\$

Simplification

9 \$\frac(1)(6) \times \frac(2)(3) / \frac(1)(6) \times \frac(2)(3) + \frac(5)(6) \times (1)(3)\$

After simplification we get

\$\frac(2)(7)\$

Answer

A